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Electricity of a Toaster

  1. Jan 16, 2011 #1
    1. The problem statement, all variables and given/known data
    A toaster, which uses 6 alkaline D batteries, must produce at least 900 W of thermal energy. D cell batteries have a capacity of about 5 Ah. How long can the batteries run the toaster?

    2. Relevant equations

    3. The attempt at a solution
    Well I know that Power is (J/s) and that ah can be converted to Coloumbs. However when I solve out 900(J/s)/10,800C, I get .00833V/s. So how do I finish this problem and solve for time? I feel like I'm close to the answer, but I can't piece together the last step. I would greatly appreciate any help. Thanks!
  2. jcsd
  3. Jan 16, 2011 #2


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    You've been provided with a total amount of battery charge. To figure out how long that charge will last, you need to know the rate at which that charge is "used up." So answer me the following questions:

    1. What is the quantity that is a measure of the rate of flow of charge called?

    2. How can you compute the quantity in part 1 from the power consumption, which is given (hint: you'll need to know something about 6 D cell batteries in series).
    Last edited: Jan 16, 2011
  4. Jan 17, 2011 #3
    1. The rate of flow is the current, is it not? This is the change in charge over the change in time.
    2. The D cells in a series all have the same current, but add together to give the total voltage and resistance. After looking at the units attached to the different quantities, do I need to use resistance at all? Or current? Sorry, I feel like all the pieces are here, I just don't know how to fit them together.
  5. Jan 17, 2011 #4


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    For #1, yes, current is what I meant.

    For #2, let me put it this way. You know the power. You need to know the current. What is the relationship between the power consumed by the load and the current flowing across the load? Hint: how much energy is lost per unit of charge in going across the load and what is this called?
  6. Jan 17, 2011 #5
    I think this makes sense now. Are you referring to the fact that Resistance is equal to potential difference/current?
  7. Jan 17, 2011 #6
    Would it be 30 hours? As in 900W/30AH=30H? Or am I missing it here?
  8. Jan 17, 2011 #7


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    That's true, but you can't figure out the resistance, because you don't know the current. And if you figure out the current, you don't *need* the resistance. Exactly how to find the current is what I have been trying to hint at for the last several posts. Let me be clearer about what I mean:

    P = VI

    This is the relationship between power and current. With what I said in the next quote, I was trying to steer you towards this relationship by explaining why this relationship is true:

    In other words, since the load is connected across the battery, the voltage drop across the load is equal to the battery terminal voltage. This voltage, or potential difference, is the energy lost per unit of charge moving across the load. The current, is the rate at which charge flows across the load. Hence, the product of the two is the rate at which the energy of the charges is consumed by the load (dissipated as heat). In other words, it is the power consumed. THAT is why power = current*voltage.

    Now, a D-cell battery has a voltage of 1.5 V. So, six of them in series would have a total voltage across them of 9 V (this is what I meant in my first post when I said that you would need to know something about D-cell batteries). Now you can calculate the current from I = P/V.

    Once you know the current, you can combine it in a certain way with the total amount of charge to figure out how long that charge will last. (Think about how the units have to work out).
  9. Jan 17, 2011 #8
    Many thanks for this, it has made the problem much more clear. Using the formula, I obtained I=100A, which is the necessary current that the batteries must supply to the toaster. Since each battery has a capacitance of 5ah, I believe I need to divide the current by the capacitance. This would yield 100A/5ah, which would mean that the batteries could supply 20 hours of power to the toaster.

    One lingering question though (assuming I did this correctly that is), why wouldn't I multiply the batteries capacitance together and then divide by that number? Thanks so much, you've been a great help!
  10. Jan 17, 2011 #9
    Wait after further reflection I think I've done this backwards. I should have divided the 5AH by the 100A, giving me .05H and then multiplied this by 6 to get .3 or 18 minutes. Is this the right way?
  11. Jan 17, 2011 #10


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    I'm afraid that there are numerous problems with this:

    Yes, that is correct.

    Be careful! This quantity is NOT capacitance. Capacitance (symbol: C) is something totally different that is measured in units of farads (F). 5 Ah is an amount of CHARGE (symbol: Q). Charge is measured in coulombs (C). You know that 5 Ah is a charge, because it has dimensions of current*time. Unfortunately, battery manufacturers often use the word "battery capacity" to describe the "total amount of charge that a battery can store." But this has nothing to do with capacitANCE. Once more for emphasis: a battery's capacity is the total amount of CHARGE it can store.


    [tex] \frac{100~\textrm{A}}{5~\textrm{Ah}} = 20~\textrm{h}^{-1} [/tex]

    Remember how I said that you would know which was the right way to combine the charge and the current because of how the units would have to work out. Well, you got an answer in units of hours-1, so obviously dividing the current by the charge was not the right way to do it!

    Even without paying attention to the units, it should be obvious that this answer can't be right. Intuitively, does it make sense that if you draw 100 A (which is a CRAZY amount of current), that 6 small batteries can last almost a full day? It doesn't make any sense, especially since 5 amp-hours of charge means that if you draw 5 A from the battery, it will last 1 hour (and if you draw 1 A from the battery, it will last 5 hours). So HOW could it possibly last 20 hours at 100 A?

    Although I'm not totally sure what you are asking here, I will try to answer. The reason why the total amount of charge from each battery can't be ADDED together to give you 30 Ah of total "capacity" (which is NOT capacitance) is because these batteries are in series. IF they were in parallel, then each battery could contribute a smaller amount of current to make up the total current. Therefore, to get 100 A total, it would suffice to draw (100/6) A from each battery, and the batteries would last 6 times longer than a single one. HOWEVER, since the batteries are in series, they can't contribute separate, different currents to the total. The total current flows through all of them, meaning that each battery has to provide the full 100 A. All batteries are drained at 100 A simultaneously. Therefore, you still have only 5 Ah of charge available to you, not 30 Ah.
  12. Jan 18, 2011 #11
    Thanks I understand that I messed up. Did you not see my second post though? I realized I needed to decide the 5AH by the 100A which yielded 18 min.
  13. Jan 18, 2011 #12


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    No I didn't see that post until now. Good correction, but you shouldn't multiply your final answer by 6 (see the last paragraph of my last post for an explanation).
  14. Jan 18, 2011 #13
    Right I figured that after your most recent explaination. I think I'm good now. Thanks so much. You were incredibly helpful.
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